Question: A jar contains $4$ red marbles, $4$ green marbles, and $5$ blue marbles. If we choose a marble, then another marble without putting the first one back in the jar, what is the probability that the first marble will be blue and the second will be green?
Answer: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened. In this case, event A is picking a blue marble and leaving it out. Event B is picking a green marble. Let's take the events one at at time. What is the probability that the first marble chosen will be blue? There are $5$ blue marbles, and $13$ total, so the probability we will pick a blue marble is $\dfrac{5} {13}$. After we take out the first marble, we don't put it back in, so there are only $12$ marbles left. Since the first marble was blue, there are still $4$ green marbles left. So, the probability of picking a green marble after taking out a blue marble is $\dfrac{4} {12}$. Therefore, the probability of picking a blue marble, then a green marble is $\dfrac{5}{13} \cdot \dfrac{4}{12} = \dfrac{20}{156} = \dfrac{5}{39}$